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March 2, 2015

March 2, 2015

Posted by **Jon** on Wednesday, February 20, 2008 at 11:55am.

A)1 real root, 1 imaginary root

B)2 real roots, 2 imaginary roots

C)2 real roots

D)4 real roots

I went with A

sqrt x^2= x (which would be the imaginary) and sqrt 625= 25 (the real)

- Math -
**Reiny**, Wednesday, February 20, 2008 at 12:09pmHow did you get that answer???

This is a straightforward difference of squares setup

(x+25)(x-25) = 0

so x = ± 25

2 real roots, so C

- Math(I see that now) -
**Jon**, Wednesday, February 20, 2008 at 12:14pmI made it harder than it was thanks. But I just thought that x was imaginary

- Math -
**Damon**, Wednesday, February 20, 2008 at 1:39pmBy the way, if you get one complex or imaginary root, you get two. They come in pairs, like male and female twins. One has +imaginary number, the other has -the same imaginary number. They are called complex conjugates. a + b i and a - b i and a can be zero.

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